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Graph/3f51d5b1-20e5-425a-92fa-4f0162188c51

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HashMap Lookup

3f51d5b1-20e5-425a-92fa-4f0162188c51

HashMap Lookup

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Updated5/1/2026

PageRank0.0000

Effectiveness0.000

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4 linked nodes

PERTAIN_TOProblem

my confusion arises from many explanations of the process as "finding the bucket and immediately running equals() method on all entries" — When running get(Object key) on an instance of a HashMap in Java. Tension: why would you run equals() if you don't even know if the hashes are the same? Outcome: As one of the people commenting mentioned, this question only concerns this specific implementation.

PERTAIN_TOSolution

the correct bucket is first accessed by calculating the following formula: (n-1) & hash — when running get(Object key) on an instance of a HashMap in Java. Outcome: If the hashes are equal, then the equals() method runs to make sure it's the same.

PERTAIN_TOProblem

Both the existingESItems and newItems are converted into sorted sets (SortedSet). Tension: even though the lists contain different values. Outcome: the code isn't behaving as expected.

PERTAIN_TOSolution

You can use CollectionUtils.isEqualCollection() from Apache — In case you want to use custom `Comparator`. Tension: Approach with using `Set` is not good.

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